Question

The solubility product of $BaC{l}_2$ is $3.2\times {10}^{-9}mo{l}^3{L}^{-3}$. What will be its solubility in mol$L^{-1}$?

• A. $4\times {10}^{-3}$
• B. $3.2\times {10}^{-9}$
• C. $1\times {10}^{-3}$
• D. $1\times {10}^{-9}$

Answer 1

Answer: (C)

$1\times {10}^{-3}$

$BaC{l}_{\rightleftharpoons }B{a}^{2+}+2C{l}^{-1}$
$K_{sp}=\left[B{a}^{2+}\right][C{l}^{-}{]}^2=x\times (2x{)}^2=4x^3$
$4x^3=3.2\times {10}^{-9}$
$\Rightarrow x=9.28\times {10}^{-4}=0.928\times {10}^{-3}=1\times {10}^{-3}$